You have found the following ages (in years) of all 6 lions at your local zoo: $ 7,\enspace 8,\enspace 5,\enspace 1,\enspace 6,\enspace 2$ What is the average age of the lions at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 6 lions at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{7 + 8 + 5 + 1 + 6 + 2}{{6}} = {4.8\text{ years old}} $ Find the squared deviations from the mean for each lion. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $7$ years $2.2$ years $4.84$ years $^2$ $8$ years $3.2$ years $10.24$ years $^2$ $5$ years $0.2$ years $0.04$ years $^2$ $1$ year $-3.8$ years $14.44$ years $^2$ $6$ years $1.2$ years $1.44$ years $^2$ $2$ years $-2.8$ years $7.84$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{4.84} + {10.24} + {0.04} + {14.44} + {1.44} + {7.84}} {{6}} $ $ {\sigma^2} = \dfrac{{38.84}}{{6}} = {6.47\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{6.47\text{ years}^2}} = {2.5\text{ years}} $ The average lion at the zoo is 4.8 years old. There is a standard deviation of 2.5 years.